Fundamental of Control Systems
– Steady State Error
Lecturer:
Dr. Wahidin Wahab M.Sc.
Aries Subiantoro, ST. MSc.
Electrical Engineering Department
University of Indonesia
2
Steady State Error
‹ How
well can a system track a standard input
¾step, ramp, parabola
¾based on final value theorem of Laplace
Transforms
Steady state value of e( t ) is
e(∞) = lim {e( t )} = lim {sE ( s)}
t →∞
s→ 0
Table 7.1
Test waveforms
for evaluating
steady-state
errors of
position control
systems
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.2
Steady-state error:
a. step input;
b. ramp input
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
5
Steady State Error
‹ Consider
a unity feedback system
¾note that any closed loop system may be
reduced to a unity gain system
R(s)
+
E(s)
-
C(s)
G(s)
6
Steady State Error
R(s)
+
E(s)
-
C(s)
G(s)
Steady state error ess = lim {sE ( s)}
s→ 0
C ( s ) E ( s) + R ( s )
G ( s)
=
=
1 + G ( s)
R ( s)
R ( s)
1
E ( s)
G ( s)
→
=
−1=
1 + G ( s)
R ( s) 1 + G ( s )
7
Steady State Error
R(s)
+
E(s)
-
G(s)
C(s)
E ( s)
1
1
=
→ E ( s) = R ( s ) ⋅
R ( s) 1 + G ( s )
1 + G ( s)
⎧ sR ( s) ⎫
∴ ess = lim ⎨
⎬
s→ 0⎩1 + G ( s) ⎭
8
Error Constants
‹ Error
constants give figure of merit for
steady state error
‹ The bigger the error constant, the better
the steady state error
‹ tracking of input signal defined in terms
of response to standard inputs
¾step (position), ramp(velocity),
parabola(acceleration)
9
System types
‹ Systems
classified into how many open
loop poles lie at the origin
‹ e.g. type 0 open loop systems have no
poles at the origin, type 1 systems have
one pole at the origin, and so on
10
Static Position Error Constant
The steady state error of a unity feedback system for a
unit step input is
1
1⎫
1
⎧
ess = lim ⎨s ⋅
⋅ ⎬=
s → 0 ⎩ 1 + G ( s) s ⎭ 1 + G ( 0 )
Define static position error constant
K p = lim {G ( s)} = G ( 0)
s→ 0
11
Static Position Error Constant
1
→ ess =
1 + Kp
Note for a type 0 system K p is finite, → ess is finite
for a type 1 or higher system it is infinite
→ ess is zero
If zero steady state error is required for a step input
a first or higher order system is needed.
12
Static Velocity Error Constant
Steady state response fo system to unit ramp input
1
1⎫
⎧ 1 ⎫
⎧
ess = lim ⎨s ⋅
⋅ 2 ⎬ = lim ⎨
⎬
s→ 0⎩ 1 + G ( s) s ⎭ s→ 0⎩ sG ( s) ⎭
Define static velocity error constant
K v = lim {sG ( s)}
s→ 0
1
→ ess =
Kv
13
Static Velocity Error Constant
For a type 0 system K v = 0 → ess is infinite
Type 1 system K v is finite → ess is finite
Type 2 or higher system K v is infinite → ess is zero
Static Acceleration Error
Constant
Steady state response fo system to unit parabola input
⎧ 1 ⎫
1
1⎫
⎧
⋅ 3 ⎬ = lim ⎨ 2
ess = lim ⎨s ⋅
⎬
s → 0 ⎩ 1 + G ( s) s ⎭ s → 0 ⎩ s G ( s ) ⎭
Define static acceleration error constant
{
s→ 0
}
K a = lim s G ( s)
1
→ ess =
Ka
2
14
Static Acceleration Error
Constant
For a system types 0 &1 K a = 0 → ess is infinite
Type 2 system K a is finite → ess is finite
Type 3 or higher system K v is infinite → ess is zero
15
Figure 7.5
Feedback
control system for
Example 7.2
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.6
Feedback
control system for
Example 7.3
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.8
Feedback control
system for defining
system type
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.10
Feedback
control system
for Example 7.6
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Table 7.2
Relationships between input, system type, static error
constants, and steady-state errors
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Steady State Error - A
Summary
SYSTEM TYPE Step Input
1
0
1 + Kp
1
2
0
0
Ramp Input
∞
21
Parabolic Input
∞
1
Kv
∞
0
1
Ka
Figure 7.11
Feedback control
system showing
disturbance
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.13
Feedback control system for
Example 7.7
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.14
System for skill-Assessment
Exercise 7.4
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.15
Forming an
equivalent
unity feedback
system from a
general nonunity
feedback system
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.16
Nonunity feedback
control system for
Example 7.8
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.17
Nonunity feedback
control system with
disturbance
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Figure 7.18
Nonunity feedback
system for
Skill-Assessment
Exercise 7.5
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
29
Example 1
‹ Taken
from Ogata, p287
‹ Consider liquid level control system.
Determine the steady state effect of
disturbance of size D0 if proportional
control is used and alternatively if
integral control is used.
30
Example 1
‹
Block Diagram
D(s)
X(s)
+
E(s)
-
controller
Gc(s)
G(s)
G(s)
++
Disturbance
H(s)
31
Example 1
R
G ( s) =
,
RCs + 1
⎧K p Proportional control
⎪
Gc ( s) = ⎨ K
Integral control
⎪⎩ s
D0
Disturbance D ( s) =
s
32
Example 1
Taking variation in setpoint as zero X( s) = 0
K pR
R
→ H ( s) =
E ( s) +
D ( s)
RCs + 1
RCs + 1
K pR
R
∴ E ( s) = − H ( s ) = −
E ( s) −
D ( s)
RCs + 1
RCs + 1
33
Example 1
R
→ E ( s) = −
D ( s)
RCs + 1 + K p R
D0
D ( s) =
s
D0
R
→ E ( s) = −
⋅
RCs + 1 + K p R s
34
Example 1
⎞
⎛
⎟
RD0 ⎜
RD0 1
1
⎟−
⎜
→ E ( s) =
⋅
1 + K pR⎟ 1 + K pR s
1 + K pR ⎜
s+
⎝
RC ⎠
RD0
e( ∞ ) = lim {sE ( s)} = −
s→ 0
1 + K pR
To get good steady state error need high value of Kp
35
Example 1
K
if controller is integral Gc ( s) =
s
Rs
E ( s) = −
D
s
(
)
RCs2 + s + KR
D0 ⎫
Rs
⎧
=
0
→ e(∞) = lim {sE ( s)} = lim ⎨− s
⎬
s→ 0
s→ 0 ⎩
RCs 2 + s + KR s ⎭
Integrator eliminates steady state error due to
step disturbance
Homeworks
Ogata Chapter
Nise chapter 7: 4, 12,16, 35, 38, 45